Problem: Evaluate − 64 − − − − √ 3 \sqrt[\leftroot{-2}\uproot{2}\scriptstyle 3]{-64}.
Solution: What number times itself and then times itself again equals $-64$ ? ${-4} \cdot {-4} \cdot {-4} = -64$ − 64 − − − − √ 3 = − 4 \sqrt[\leftroot{-2}\uproot{2}\scriptstyle 3]{-64}=-4